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HCP distributution in a 2NT opener


sartaj1

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This sounds about right. Everything else being equal, you're more likely to have a number of points close to ten. You can see from experience that hands with 12 points, say, are VERY common, whereas hands with 0 points or 20 points are few and far between. So not all numbers of points are created equal, it follows something vaguely resembling a bell-shape curve with 10 (the average) being most common.

 

So it won't surprise you that the probabilities of holding different numbers of points look like:

 

20: 0.64% of hands

21: 0.38% of hands

22: 0.21% of hands

 

For comparison:

 

9: 9.36% of hands

10: 9.41% of hands

11: 8.94% of hands

 

Okay, let's suppose we have a 20-22 notrump. If we could open any 20-22 point hand in this range, the probability of exactly 20 points is equal to:

 

Pr[20 points given 20-22] = Pr[20 points] / Pr[20 or 21 or 22] = 0.64% / 1.23% or 52%.

 

Of course, this may be adjusted slightly by the knowledge that opener must have a balanced hand, but this honestly shouldn't be a big effect.

 

P.S. Thanks to the ZAR book for the hcp probabilities.

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That calculation is right.

If you want to get picky, and define a 2NT opener as 20-22 HCP with distributions

- any 4432

- any 4333

- any 5332

- any 6322 with a 6-card minor

- any 5422 without both majors

 

(which is about how I play it, obviously you can be more or less strict, and some people allow singleton honours and/or 4441s)

 

then 20 HCP = 0.41%, 21 = 0.24%, 22=0.14%

and the relative frequencies are unchanged.

 

In my case, the percentage of 20 counts is a little higher, because the range is slightly better put as:

 

20-22 4432

20-22 4333

20-21 5332 / 6332 6-card minor/ 5422 without both majors

 

which makes a 20 count about 56%, and 20-21 about 89%

 

(I seldom upgrade 19-counts)

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